(a)
If \(\mid G \mid =p^2q\) , then \(\mid \operatorname{Cent}(G) \mid = q+2\)  or \(G=A_4\) .
(b)
If \(\mid G \mid =pq^2\) , then \(\mid \operatorname{Cent}(G) \mid = q+2\)  or \(q^2+2\) .

a) If \(\mid G \mid =12\) , then \(G\cong A_4\) , \(T= \langle x, y \mid x^6=1, y^2=x^3, y^{-1}xy=x^{-1} \rangle \)  or \(D_{12}\) .
It is easy to see that \(\mid \operatorname{Cent}(A_4) \mid = 6\) , \(\mid \operatorname{Cent}(T) \mid = 5\)  and \(\mid \operatorname{Cent}(D_{12}) \mid = 5\) .
Next suppose \(\mid G \mid >12\) . Clearly, we have \(\mid Z(G) \mid =1\)  or \(p\) .
Now, if \(\mid Z(G) \mid =p\) , then \(\mid \frac{G}{Z(G)} \mid =pq\)  and hence by [1]}, we have \(\mid \operatorname{Cent}(G) \mid = q+2\) .
On the otherhand if \(\mid Z(G) \mid =1\) , then for any Sylow subgroup \(S\)  of \(G\)  and any \(x (\ne 1) \in S\) , we have \(C(x)=S\) . Hence \(\mid \operatorname{Cent}(G) \mid = n_p+n_q+1=q+2\) .
b) Clearly, we have \(\mid Z(G) \mid =1\)  or \(q\) .
Now, if \(\mid Z(G) \mid =q\) , then \(\mid \frac{G}{Z(G)} \mid =pq\)  and hence by [1]}, we have \(\mid \operatorname{Cent}(G) \mid = q+2\) .
On the otherhand if \(\mid Z(G) \mid =1\) , then for any Sylow subgroup \(S\)  of \(G\)  and any \(x (\ne 1) \in S\)  we have \(C(x)=S\) . Hence \(\mid \operatorname{Cent}(G) \mid = n_p+n_q+1=q^2+2\) .
The following lemma will be used in proving the next two results.
Lemma 2.9 
Let \(G\)  be a finite CA-group. If \(\frac{G}{Z(G)}\)  is abelian, then \(\frac{G}{Z(G)}\)  is elementary abelian.

We have \(\lbrace \frac{C(x)}{Z(G)} / x \in G \setminus Z(G)\rbrace \)  is a partition of \(\frac{G}{Z(G)}\)  (see [3]}). Therefore by [4]}, we have \(\frac{G}{Z(G)}\)  is elementary abelian.
Let \(p\)  and \(q\)  be two distinct primes. It may be mentioned here that in [5]}, the authors characterised non-abelian capable groups of order \(p^2q\)  using many technical and sophisticated tools. In the following result, we have characterised any capable group of order \(p^2q\) . We present a very elementary proof of this result using basic group theory.
Remark 2.10 
Let \(p<q\)  be primes. By [6]}, the only non-abelian group of order \(p^2q\)  such that \(S_p=C_p \times C_p\)  is \(C_p \times (C_q \rtimes C_p)\) .

Theorem 2.11 
Let \(G\)  be any group and \(p<q\)  be primes.
(a)
Suppose \(\mid G \mid = pq^2\) . Then \(G\)  is a capable group if and only if \(\mid Z(G) \mid =1\) .
(b)
Suppose \(\mid G \mid = p^2q\) . Then \(G\)  is a capable group if and only if \(\mid Z(G) \mid =1\)  or \(G \cong C_p \times (C_q \rtimes C_p)\) .

By Remark REF , \(G\)  is a CA-group. Therefore in view of Lemma REF , \(G\)  is non-abelian. Now,
a) Suppose \(G\)  is a capable group of order \(pq^2\) . Then there exists a group \(H\)  such that \(G \cong \frac{H}{Z(H)}\) . By Remark REF , we have both \(G\)  and \(H\)  are CA-groups. Now, suppose \(\mid Z(\frac{H}{Z(H)}) \mid =q\) . Note that \(\frac{H}{Z(H)}\)  has exactly one centralizer of order \(q^2\) . Let \(C(sZ(H))\)  and \(C(tZ(H))\)  be two distinct centralizers of \(\frac{H}{Z(H)}\)  of order \(pq\) . Then \(C(sZ(H))=\langle uZ(H) \rangle =\frac{C(u)}{Z(H)}\)  and \(C(tZ(H))=\langle vZ(G) \rangle =\frac{C(v)}{Z(H)}\) . It now follows that \(\mid \frac{H}{C(u)}\mid = p= \mid \frac{H}{C(v)}\mid \) .
But then \(\mid \frac{H}{Z(H)} \mid = \mid \frac{H}{{C(u) \cap C(v)}} \mid \le p^2\) , which is a contradiction. Therefore \(\mid Z(\frac{H}{Z(H)}) \mid =1\) . Converse is trivial.
b) Suppose \(G\)  is a capable group of order \(p^2q\) . Then there exists a group \(H\)  such that \(G \cong \frac{H}{Z(H)}\) . Now, suppose \(\mid Z(\frac{H}{Z(H)}) \mid =p\) . Let \(C(sZ(H))\)  and \(C(tZ(H))\)  be two centralizers of \(\frac{H}{Z(H)}\)  of order \(p^2\)  and \(pq\)  respectively. If \(S_p=C_{p^2}\) , then \(C(sZ(H))=\langle uZ(H) \rangle =\frac{C(u)}{Z(H)}\) . Also we have \(C(tZ(H))=\langle vZ(G) \rangle =\frac{C(v)}{Z(H)}\) . It now follows that \(\mid \frac{H}{C(u)}\mid = q\)  and \(\mid \frac{H}{C(v)}\mid =p\) .
But then \(\mid \frac{H}{Z(H)} \mid = \mid \frac{H}{{C(u) \cap C(v)}} \mid \le pq\) , which is a contradiction. Therefore \(S_p = C_p \times C_p\)  and hence \(G \cong C_p \times (C_q \rtimes C_p)\)  by Remark REF .
Conversely, if \(\mid Z(G) \mid =1\) , then \(G\)  is capable. On the otherhand, from [7]}, consider any group \(H\)  of order \(p^3q\)  given by \(H=\langle a, b, c, d \mid a^p=b^p=c^p=d^q=1, ab=ba, ac=ca, ad=da, bd=db, c^{-1}bc=ab, c^{-1}dc=d^i \rangle \) , where \(q \equiv 1 \;(mod \;p)\)  and \(i^p \equiv 1 \;(mod \; q)\) . Then \(\frac{H}{Z(H)}\)  is of order \(p^2q\)  with \(p\) -Sylow subgroup \(S_p = C_p \times C_p\) . Therefore \( \frac{H}{Z(H)} \cong C_p \times (C_q \rtimes C_p)\)  by Remark REF .
Now we compute \(\mid Cent(G) \mid \)  for groups whose central factor is of order \(p^2q\)  for any primes \(p, q\) . Note that the group \(T= \langle x, y \mid x^6=1, y^2=x^3, y^{-1}xy=x^{-1} \rangle \)  has non-trivial center and cyclic 2-Sylow subgroups. Therefore \(T\)  is not capable by Theorem REF .
Proposition 2.12 
Let \(G\)  be a finite group and \(p<q\)  be primes.
(a)
If \(\mid \frac{G}{Z(G)} \mid =12\) , then \(\mid \operatorname{Cent}(G) \mid = 6\)  or 8. Otherwise,
(b)
If \(\mid \frac{G}{Z(G)} \mid =p^2q\) , then \(\mid \operatorname{Cent}(G) \mid = pq+2\)  or \(q+2\) .
(c)
If \(\mid \frac{G}{Z(G)} \mid =pq^2\) , then \(\mid \operatorname{Cent}(G) \mid = q^2+2\)  or \(q^2+q+2\) .

In view of Remark REF  and Lemma REF , \(\frac{G}{Z(G)}\)  is non-abelian. Now,
a) We have \(\frac{G}{Z(G)} \cong A_4\)  or \(D_{12}\) . If \(\frac{G}{Z(G)} \cong A_4\) , then \(G\)  has exactly 4 centralizers of index 4 and these will absorb exactly \(\frac{2\mid G \mid }{3}\)  non-central elements of \(G\) , noting that \(G\)  is a CA-group by Remark REF . Let \(k\)  be the number of centralizers produced by the remaining \(\frac{\mid G \mid }{3}\)  elements of \(G\) . Now, if \(G\)  has a centralizer of index 3, then \(k=1\) . Suppose \(G\)  has no centralizer of index 3. Then \(\frac{\mid G \mid }{3}=k(\frac{\mid G \mid }{6}-\frac{\mid G \mid }{12})+ \frac{\mid G \mid }{12}\)  and consequently, \(k=3\) . Hence \(\mid \operatorname{Cent}(G) \mid =6\)  or 8. Again, if \(\frac{G}{Z(G)} \cong D_{12}\) , then by [1]}, \(\mid \operatorname{Cent}(G) \mid =8\) .
b) Clearly \(\mid Z(\frac{G}{Z(G)}) \mid =1\)  or \(p\) . Now, If \(\mid Z(\frac{G}{Z(G)}) \mid =p\) , then by Theorem REF , \(\frac{G}{Z(G)} \cong C_{pq} {\rtimes }_\theta C_p\)  and so \(\mid \operatorname{Cent}(G) \mid = pq+2\)  by [1]}.
Again if \(\mid Z(\frac{G}{Z(G)}) \mid =1\) , then we have \(C(xZ(G))=S_q\)  for any \(xZ(G) (\ne 1) \in S_q\) . Therefore by [10]}, \(\frac{G}{Z(G)}\)  is a Frobenius group with Frobenius kernel \(\frac{C(x)}{Z(G)}=C(xZ(G))\)  and cyclic Frobenius complement \(\langle yZ(G)\rangle = \frac{C(y)}{Z(G)}\)  of order \(p^2\) . Moreover by Remark REF , \(C(y)\)  is abelian . Therefore by [11]}, we have \(\mid \operatorname{Cent}(G) \mid = q+2\) .
c) By Theorem REF , we have \(\mid Z(\frac{G}{Z(G)}) \mid =1\) . Therefore \(\frac{G}{Z(G)}\)  is a Frobenius group with Frobenius kernel \(\frac{K}{Z(G)}\)  and Frobenius complement \(\frac{H}{Z(G)}\)  of order \(p\) . Now, using [11]}, we have \(\mid \operatorname{Cent}(G) \mid = q^2+2\)  or \(q^2+q+2\)  (noting that if \(Z(G)=Z(K)\) , then \(\mid \operatorname{Cent}(K) \mid =q+2\)  by [1]}).
Finally, we consider the groups of order \(p^3\) .
Remark 2.13 
Let \(G\)  be a group of order \(p^3\) . If \(G\)  is non-abelian then by [14]}, \(G\)  is capable if and only if \(G \cong D_8\)  or a \(p\)  group of exponent \(p\) . It may be noted that \(\frac{G}{Z(G)} \cong C_2 \times C_2 \times C_2\)  for the group \(G=Small Group (64, 60)\)  in [15]} (it is also pointed in [3]}). Again, if \(p>2\) , we have \(\frac{G}{Z(G)} \cong C_p \times C_p \times C_p\)  for the group \(G\) =Group \(5.3.1\)  in [17]}. Therefore, if \(G\)  is abelian then in view of Lemma REF , \(G\)  is capable if and only if \(G \cong C_p \times C_p \times C_p\) .

It is easy to see that for any non-abelian group \(G\)  of order \(p^3\) , we have \(\mid \operatorname{Cent}(G) \mid = p+2\) . We now compute \(\mid \operatorname{Cent}(G) \mid \)  for any group \(G\)  (finite or infinite) whose central factor is of order \(p^3\) . This extends [18]} and [19]}. It may be mentioned here that \(\mid \operatorname{Cent}(G) \mid =\omega (G)+1\)  if and only if \(G\)  is a CA-group (see arguments of [3]}), where \(\omega (G)\)  is the size of a maximal set of pairwise non-commuting elements of \(G\) .
Proposition 2.14 
Let \(G\)  be any group (finite or infinite) and \(p\)  be a prime. If \(\mid \frac{G}{Z(G)} \mid =p^3\) , then \(\mid \operatorname{Cent}(G) \mid = \omega (G)+1=p^2+p+2\)  or \(p^2+2\) .

By Remark REF , \(C(x) \cap C(y) = Z(G)\)  for any \(x,y \in G \setminus Z(G), xy \ne yx\) . Now, suppose \(G\)  has no centralizer of index \(p\) . Then each proper centralizer of \(G\)  will contain exactly \(p\)  distinct right coset of \(Z(G)\) . Therefore \(\mid \operatorname{Cent}(G) \mid = p^2+p+2\) .
Next, suppose \(G\)  has a centralizer \(C(x)\)  of index \(p\) . Then \(C(x)\)  will contain exactly \(p^2\)  distinct right cosets of \(Z(G)\) . Therefore the number of right cosets of \(Z(G)\)  (other than \(Z(G)\) ) left for the remaining centralizers is \(p^3-p^2\) . Clearly, in view of Second Isomorphism Theorem, \(G\)  cannot have another centralizer of index \(p\)  (noting that in the present scenario \(C(x) \lhd G\)  ) and consequently, any proper centralizer other than \(C(x)\)  will contain exactly \(p\)  distinct right cosets of \(Z(G)\) . Hence \(\mid \operatorname{Cent}(G) \mid = p^2+2\) .
